Определен интеграл

[krot]

∫(dx/(x^2-2*x*cosa+1) 0<a<∏ from -1 to 1

[mp3]

[tex]=\int_{-1}^{ 1}\frac{dx}{ x^{2} -2x.cos\alpha +1+cos^{2}\alpha -cos^{2}\alpha} =[/tex]
[tex]= \int_{-1}^{ 1}\frac{dx}{ (x -cos\alpha)^{2} +1-cos^{2}\alpha}=[/tex]
[tex]= \int_{-1}^{ 1}\frac{dx}{ (x -cos\alpha)^{2} +sin^{2}\alpha}=[/tex]
[tex]= \frac{1}{ sin^{2}\alpha } \int_{-1}^{ 1}\frac{dx}{ (\frac{x -cos\alpha}{ sin\alpha} )^{2} +1}=[/tex]
[tex]= \frac{sin\alpha}{ sin^{2}\alpha } \int_{-1}^{ 1}\frac{\frac{dx}{ sin\alpha} }{ (\frac{x -cos\alpha}{ sin\alpha} )^{2} +1}=[/tex]
[tex]= \frac{sin\alpha}{ sin^{2}\alpha } \int_{-1}^{ 1}\frac{ d\frac{x}{ sin\alpha} }{ (\frac{x -cos\alpha}{ sin\alpha} )^{2} +1}=[/tex]
[tex]= \frac{1}{ sin\alpha } \int_{-1}^{ 1}\frac{d \frac{x-cos\alpha}{ sin\alpha} }{ 1+ (\frac{x -cos\alpha}{ sin\alpha} )^{2} }=[/tex]
[tex]= \frac{1}{ sin\alpha} .arctg(\frac{x-cos\alpha}{ sin\alpha} ) |\frac{1}{-1 } =[/tex]